3.60 \(\int \frac{(1+x)^2}{x^3 \sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=51 \[ -\frac{2 \sqrt{1-x^2}}{x}-\frac{\sqrt{1-x^2}}{2 x^2}-\frac{3}{2} \tanh ^{-1}\left (\sqrt{1-x^2}\right ) \]

[Out]

-Sqrt[1 - x^2]/(2*x^2) - (2*Sqrt[1 - x^2])/x - (3*ArcTanh[Sqrt[1 - x^2]])/2

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Rubi [A]  time = 0.0614837, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1807, 807, 266, 63, 206} \[ -\frac{2 \sqrt{1-x^2}}{x}-\frac{\sqrt{1-x^2}}{2 x^2}-\frac{3}{2} \tanh ^{-1}\left (\sqrt{1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^2/(x^3*Sqrt[1 - x^2]),x]

[Out]

-Sqrt[1 - x^2]/(2*x^2) - (2*Sqrt[1 - x^2])/x - (3*ArcTanh[Sqrt[1 - x^2]])/2

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(1+x)^2}{x^3 \sqrt{1-x^2}} \, dx &=-\frac{\sqrt{1-x^2}}{2 x^2}-\frac{1}{2} \int \frac{-4-3 x}{x^2 \sqrt{1-x^2}} \, dx\\ &=-\frac{\sqrt{1-x^2}}{2 x^2}-\frac{2 \sqrt{1-x^2}}{x}+\frac{3}{2} \int \frac{1}{x \sqrt{1-x^2}} \, dx\\ &=-\frac{\sqrt{1-x^2}}{2 x^2}-\frac{2 \sqrt{1-x^2}}{x}+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1-x^2}}{2 x^2}-\frac{2 \sqrt{1-x^2}}{x}-\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{1-x^2}\right )\\ &=-\frac{\sqrt{1-x^2}}{2 x^2}-\frac{2 \sqrt{1-x^2}}{x}-\frac{3}{2} \tanh ^{-1}\left (\sqrt{1-x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0195676, size = 40, normalized size = 0.78 \[ -\frac{\sqrt{1-x^2} (4 x+1)}{2 x^2}-\frac{3}{2} \tanh ^{-1}\left (\sqrt{1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^2/(x^3*Sqrt[1 - x^2]),x]

[Out]

-((1 + 4*x)*Sqrt[1 - x^2])/(2*x^2) - (3*ArcTanh[Sqrt[1 - x^2]])/2

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Maple [A]  time = 0.052, size = 42, normalized size = 0.8 \begin{align*} -{\frac{3}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ) }-{\frac{1}{2\,{x}^{2}}\sqrt{-{x}^{2}+1}}-2\,{\frac{\sqrt{-{x}^{2}+1}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^2/x^3/(-x^2+1)^(1/2),x)

[Out]

-3/2*arctanh(1/(-x^2+1)^(1/2))-1/2/x^2*(-x^2+1)^(1/2)-2*(-x^2+1)^(1/2)/x

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Maxima [A]  time = 1.48397, size = 73, normalized size = 1.43 \begin{align*} -\frac{2 \, \sqrt{-x^{2} + 1}}{x} - \frac{\sqrt{-x^{2} + 1}}{2 \, x^{2}} - \frac{3}{2} \, \log \left (\frac{2 \, \sqrt{-x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^2/x^3/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-2*sqrt(-x^2 + 1)/x - 1/2*sqrt(-x^2 + 1)/x^2 - 3/2*log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))

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Fricas [A]  time = 1.98484, size = 97, normalized size = 1.9 \begin{align*} \frac{3 \, x^{2} \log \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) - \sqrt{-x^{2} + 1}{\left (4 \, x + 1\right )}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^2/x^3/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/2*(3*x^2*log((sqrt(-x^2 + 1) - 1)/x) - sqrt(-x^2 + 1)*(4*x + 1))/x^2

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Sympy [C]  time = 6.60171, size = 116, normalized size = 2.27 \begin{align*} 2 \left (\begin{cases} - \frac{i \sqrt{x^{2} - 1}}{x} & \text{for}\: \left |{x^{2}}\right | > 1 \\- \frac{\sqrt{1 - x^{2}}}{x} & \text{otherwise} \end{cases}\right ) + \begin{cases} - \frac{\operatorname{acosh}{\left (\frac{1}{x} \right )}}{2} - \frac{\sqrt{-1 + \frac{1}{x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{x^{2}}\right |} > 1 \\\frac{i \operatorname{asin}{\left (\frac{1}{x} \right )}}{2} - \frac{i}{2 x \sqrt{1 - \frac{1}{x^{2}}}} + \frac{i}{2 x^{3} \sqrt{1 - \frac{1}{x^{2}}}} & \text{otherwise} \end{cases} + \begin{cases} - \operatorname{acosh}{\left (\frac{1}{x} \right )} & \text{for}\: \frac{1}{\left |{x^{2}}\right |} > 1 \\i \operatorname{asin}{\left (\frac{1}{x} \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**2/x**3/(-x**2+1)**(1/2),x)

[Out]

2*Piecewise((-I*sqrt(x**2 - 1)/x, Abs(x**2) > 1), (-sqrt(1 - x**2)/x, True)) + Piecewise((-acosh(1/x)/2 - sqrt
(-1 + x**(-2))/(2*x), 1/Abs(x**2) > 1), (I*asin(1/x)/2 - I/(2*x*sqrt(1 - 1/x**2)) + I/(2*x**3*sqrt(1 - 1/x**2)
), True)) + Piecewise((-acosh(1/x), 1/Abs(x**2) > 1), (I*asin(1/x), True))

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Giac [B]  time = 1.1394, size = 123, normalized size = 2.41 \begin{align*} \frac{x^{2}{\left (\frac{8 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}{x} - 1\right )}}{8 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}} - \frac{\sqrt{-x^{2} + 1} - 1}{x} + \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{8 \, x^{2}} + \frac{3}{2} \, \log \left (-\frac{\sqrt{-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^2/x^3/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/8*x^2*(8*(sqrt(-x^2 + 1) - 1)/x - 1)/(sqrt(-x^2 + 1) - 1)^2 - (sqrt(-x^2 + 1) - 1)/x + 1/8*(sqrt(-x^2 + 1) -
1)^2/x^2 + 3/2*log(-(sqrt(-x^2 + 1) - 1)/abs(x))